Question

$\left[Fe\right(H_{2}O{)}_4\left(CN{)}_2\right]$ is the empirical formula of a compound which has a magnetic moment corresponding to $2\frac{2}{3}$ unpaired electrons per iron. The best possible formula of the compound is

• A. $\left[Fe\right(H_{2}O{)}_6\left]2\right[Fe\left(CN{)}_6\right]$
• B. $\left[Fe\right(H_2{O}_6\left)\right]\left[Fe\right(H_{2}O{)}_2\left(CN{)}_4\right]$
• C. $\left[Fe\right(H_{2}O{)}_4\left(CN{)}_2{]}_2\right[Fe\left(H_{2}O{)}_4\right(CN{)}_2]$
• D. None of these

Answer 1

The correct option is A $\left[Fe\right(H_{2}O{)}_6\left]2\right[Fe\left(CN{)}_6\right]$

Magnetic moment $M=\sqrt{n(n+2)}$ magneton And here $M=2\frac{2}{3}=\frac{8}{3}$ unpaired $e^{-}$ per iron.
best formula of compound is $\left[Fe\right(H_{2}O{)}_6{]}_2\left[Fe\right(CN{)}_6]$